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$2014$ points are chosen inside a cube with side $13$. Can a cube with side $1$ be found inside it so that it doesn't contain any of chosen points?

This must be a problem solved using pigeonhole principle, however I still can't see a clever way to start.

VividD
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    The first step is to compute $13 \times 13 \times 13$. Let us know when you've done that. – TonyK Sep 22 '14 at 22:19
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    @TonyK: though $12 \times 12 \times 12$ (the number of lattice points strictly inside the cube) is rather smaller, so the problem may not be quite that easy – Henry Sep 22 '14 at 22:24
  • @Henry: Yes, if you count a point on the boundary of a cube as being 'contained' in that cube, then it's not so simple. – TonyK Sep 22 '14 at 22:45

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